lim(x→0)(x^2+cosx-2)/(x^3)*ln(1+x)
=lim(x→0)(0+1-2)*(ln(1+x)/(x^3))
=lim(x→0) -(ln(1+x)/(x^3))
=im(x→0) -1/[(1+x)*(3x^2)]
=im(x→0) -1/(3x^2)
負(fù)無(wú)窮im(x→0)(x^2+2cosx-2)/(x^3)*ln(1+x)=lim(x→0)(x^2+2cosx-2)*(ln(1+x)/(x^3))把cosx，ln(1+x)展開(kāi)cosx=1-x^2/2+x^4/24……ln(1+x)=x-x^2/2+x^3/2+……lim(x→0)(x^2+2cosx-2)/[(x^3)*(ln(1+x)]=lim(x→0)(x^2+2-x^2+x^4/12-2)/(x^3)[(x-x^2/2+x^3/2)]=lim(x→0)(x^4/12)/(x^3)[(x-x^2/2+x^3/2)]=lim(x→0)(x^4/12)/(x^3)x=1/12那一步寫(xiě)錯(cuò)了，后面是對(duì)的im(x→0)(x^2+2cosx-2)/(x^3)*ln(1+x)=lim(x→0)(x^2+2cosx-2/[(x^3))(ln(1+x)]把cosx，ln(1+x)展開(kāi)cosx=1-x^2/2+x^4/24……ln(1+x)=x-x^2/2+x^3/2+……lim(x→0)(x^2+2cosx-2)/[(x^3)*(ln(1+x)]=lim(x→0)(x^2+2-x^2+x^4/12-2)/(x^3)[(x-x^2/2+x^3/2)]=lim(x→0)(x^4/12)/(x^3)[(x-x^2/2+x^3/2)]=lim(x→0)(x^4/12)/(x^3)x=1/12